If you have 20kg of water at 100°C, how much ice (in kilograms) would you have to add to that water in order to bring the temperature down to 30°C?

Question

I tried using the equation Q=mc∆T, but I keep on running into the roadblock where I can't find ∆T of the ice (the problem doesn't state the initial temperature of the ice). Any help would be appreciated.

Answer 1

-q_water = q_ice

-(mcΔT)_water = (mcΔT)_ice

-20 kg x 4.184 J/gºC x (30 - 100 ºC ) = m x 2.09 J/gºC x (30 - 0 ºC)

5858 = 62.7m

m = 5858 / 62.7

= 93.4 kg